Resolucao Beer Johnston completo

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PROBLEM 2.1 
Two forces are applied to an eye bolt fastened to a beam. Determine 
graphically the magnitude and direction of their resultant using (a) the 
parallelogram law, (b) the triangle rule. 
 
SOLUTION 
(a) 
 
 
 
 
 
 
 
 
 
(b) 
 
 
 
 
 
 
 We measure: 8.4 kNR = 
 19α = ° 
 8.4 kN=R 19°W 
1
 
 
 
PROBLEM 2.2 
The cable stays AB and AD help support pole AC. Knowing that the 
tension is 500 N in AB and 160 N in AD, determine graphically the 
magnitude and direction of the resultant of the forces exerted by the stays 
at A using (a) the parallelogram law, (b) the triangle rule. 
 
SOLUTION 
 
 
 
 
 
 
 
We measure: 51.3 , 59α β= ° = ° 
(a) 
 
 
 
 
 
 
 
(b) 
 
 
 
 
 
 
 We measure: 575 N, 67α= = °R 
 575 N=R 67°W 
2
 
 
 
 
 
PROBLEM 2.3 
Two forces P and Q are applied as shown at point A of a hook support. 
Knowing that P = 15 lb and Q = 25 lb, determine graphically the 
magnitude and direction of their resultant using (a) the parallelogram law, 
(b) the triangle rule. 
 
SOLUTION 
(a) 
 
 
 
 
 
 
 
 
 
(b) 
 
 
 
 
 
 
 
 
 
 
 We measure: 37 lb, 76α= = °R 
 37 lb=R 76°W 
3
 
 
 
 
PROBLEM 2.4 
Two forces P and Q are applied as shown at point A of a hook support. 
Knowing that P = 45 lb and Q = 15 lb, determine graphically the 
magnitude and direction of their resultant using (a) the parallelogram law, 
(b) the triangle rule. 
 
SOLUTION 
(a) 
 
 
 
 
 
 
 
 
 
(b) 
 
 
 
 
 
 
 
 
 We measure: 61.5 lb, 86.5α= = °R 
 61.5 lb=R 86.5°W 
4
 
 
 
 
 
PROBLEM 2.5 
Two control rods are attached at A to lever AB. Using trigonometry and 
knowing that the force in the left-hand rod is F1 = 120 N, determine 
(a) the required force F2 in the right-hand rod if the resultant R of the 
forces exerted by the rods on the lever is to be vertical, (b) the 
corresponding magnitude of R. 
 
SOLUTION 
 
 
 
 
 
Graphically, by the triangle law 
We measure: 2 108 NF ≅ 
 77 NR ≅ 
By trigonometry: Law of Sines 
2 120
sin sin 38 sin
F R
α β= =° 
90 28 62 , 180 62 38 80α β= ° − ° = ° = ° − ° − ° = ° 
Then: 
2 120 N
sin 62 sin 38 sin80
F R= =° ° ° 
 or (a) 2 107.6 NF = W 
 (b) 75.0 NR = W 
5
 
 
 
 
PROBLEM 2.6 
Two control rods are attached at A to lever AB. Using trigonometry and 
knowing that the force in the right-hand rod is F2 = 80 N, determine 
(a) the required force F1 in the left-hand rod if the resultant R of the 
forces exerted by the rods on the lever is to be vertical, (b) the 
corresponding magnitude of R. 
 
SOLUTION 
 
Using the Law of Sines 
1 80
sin sin 38 sin
F R
α β= =° 
90 10 80 , 180 80 38 62α β= ° − ° = ° = ° − ° − ° = ° 
Then: 
1 80 N
sin80 sin 38 sin 62
F R= =° ° ° 
 or (a) 1 89.2 NF = W 
 (b) 55.8 NR = W 
6
 
 
 
 
 
PROBLEM 2.7 
The 50-lb force is to be resolved into components along lines -a a′ and 
- .b b′ (a) Using trigonometry, determine the angle α knowing that the 
component along -a a′ is 35 lb. (b) What is the corresponding value of 
the component along - ?b b′ 
 
SOLUTION 
 
Using the triangle rule and the Law of Sines 
(a) sin sin 40
35 lb 50 lb
β °= 
sin 0.44995β = 
 26.74β = ° 
 Then: 40 180α β+ + ° = ° 
 113.3α = °W 
(b) Using the Law of Sines: 
50 lb
sin sin 40
bbF
α
′ = ° 
 71.5 lbbbF ′ = W 
7
 
 
 
 
PROBLEM 2.8 
The 50-lb force is to be resolved into components along lines -a a′ and 
- .b b′ (a) Using trigonometry, determine the angle α knowing that the 
component along -b b′ is 30 lb. (b) What is the corresponding value of 
the component along - ?a a′ 
 
SOLUTION 
 
 
 
 
 
Using the triangle rule and the Law of Sines 
(a) sin sin 40
30 lb 50 lb
α °= 
sin 0.3857α = 
 22.7α = °W 
(b) 40 180α β+ + ° = ° 
117.31β = ° 
 50 lb
sin sin 40
aaF
β
′ = ° 
 sin50 lb
sin 40
β
′
 =  ° aaF 
 69.1 lbaaF ′ = W 
8
 
 
 
 
 
PROBLEM 2.9 
To steady a sign as it is being lowered, two cables are attached to the sign 
at A. Using trigonometry and knowing that α = 25°, determine (a) the 
required magnitude of the force P if the resultant R of the two forces 
applied at A is to be vertical, (b) the corresponding magnitude of R. 
 
SOLUTION 
 
Using the triangle rule and the Law of Sines 
Have: ( )180 35 25α = ° − ° + ° 
 120= ° 
Then: 360 N
sin 35 sin120 sin 25
P R= =° ° ° 
 or (a) 489 NP = W 
 (b) 738 NR = W 
9
 
 
 
 
PROBLEM 2.10 
To steady a sign as it is being lowered, two cables are attached to the sign 
at A. Using trigonometry and knowing that the magnitude of P is 300 N, 
determine (a) the required angle α if the resultant R of the two forces 
applied at A is to be vertical, (b) the corresponding magnitude of R. 
 
SOLUTION 
 
Using the triangle rule and the Law of Sines 
(a) Have: 360 N 300 N
sin sin 35α = ° 
sin 0.68829α = 
 43.5α = °W 
(b) ( )180 35 43.5β = − ° + ° 
 101.5= ° 
 Then: 300 N
sin101.5 sin 35
R =° ° 
 or 513 NR = W 
 
10
 
 
 
 
PROBLEM 2.11 
Two forces are applied as shown to a hook support. Using trigonometry 
and knowing that the magnitude of P is 14 lb, determine (a) the required 
angle α if the resultant R of the two forces applied to the support is to be 
horizontal, (b) the corresponding magnitude of R. 
 
SOLUTION 
Using the triangle rule and the Law of Sines 
 
(a) Have: 20 lb 14 lb
sin sin 30α = ° 
 sin 0.71428α = 
 45.6α = °W 
(b) ( )180 30 45.6β = ° − ° + ° 
 104.4= ° 
 Then: 14 lb
sin104.4 sin 30
R =° ° 
 27.1 lbR = W 
11
 
 
 
 
PROBLEM 2.12 
For the hook support of Problem 2.3, using trigonometry and knowing 
that the magnitude of P is 25 lb, determine (a) the required magnitude of 
the force Q if the resultant R of the two forces applied at A is to be 
vertical, (b) the corresponding magnitude of R. 
Problem 2.3: Two forces P and Q are applied as shown at point A of a 
hook support. Knowing that P = 15 lb and Q = 25 lb, determine 
graphically the magnitude and direction of their resultant using (a) the 
parallelogram law, (b) the triangle rule. 
 
SOLUTION 
Using the triangle rule and the Law of Sines 
 
(a) Have: 25 lb
sin15 sin 30
Q =° ° 
 12.94 lbQ = W 
(b) ( )180 15 30β = ° − ° + ° 
 135= ° 
 Thus: 25 lb
sin135 sin 30
R =° ° 
sin13525 lb 35.36 lb
sin 30
R ° = = °  
 35.4 lbR = W 
12
 
 
 
 
PROBLEM 2.13 
For the hook support of Problem 2.11, determine, using trigonometry, 
(a) the magnitude and direction of the smallest force P for which the 
resultant R of the two forces applied to the support is horizontal, 
(b) the corresponding magnitude of R. 
Problem 2.11: Two forces are applied as shown to a hook support. Using 
trigonometry and knowing that the magnitude of P is 14 lb, determine 
(a) the required angle α if the resultant R of the two forces applied to the 
support is to be horizontal, (b) the corresponding magnitude of R. 
 
SOLUTION 
(a) The smallest force P will be perpendicular to R, that is, vertical 
 
( )20 lb sin 30P = ° 
 10 lb= 10 lb=P W 
(b) ( )20 lb cos30R = ° 
 17.32 lb= 17.32 lbR = W 
13
 
 
 
 
PROBLEM 2.14 
As shown in Figure P2.9, two cables are attached to a sign at A to steady 
the sign as it is being lowered. Using trigonometry, determine (a) the 
magnitude
and direction of the smallest force P for which the resultant R 
of the two forces applied at A is vertical, (b) the corresponding magnitude 
of R. 
 
SOLUTION 
We observe that force P is minimum when is 90 ,α ° that is, P is horizontal 
 
Then: (a) ( )360 N sin 35P = ° 
 or 206 N=P W 
And: (b) ( )360 N cos35R = ° 
 or 295 NR = W 
14
 
 
 
PROBLEM 2.15 
For the hook support of Problem 2.11, determine, using trigonometry, the 
magnitude and direction of the resultant of the two forces applied to the 
support knowing that P = 10 lb and α = 40°. 
Problem 2.11: Two forces are applied as shown to a hook support. Using 
trigonometry and knowing that the magnitude of P is 14 lb, determine 
(a) the required angle α if the resultant R of the two forces applied to the 
support is to be horizontal, (b) the corresponding magnitude of R. 
 
SOLUTION 
Using the force triangle and the Law of Cosines 
 
( ) ( ) ( )( )2 22 10 lb 20 lb 2 10 lb 20 lb cos110R = + − ° 
 ( ) 2100 400 400 0.342 lb = + − −  
 2636.8 lb= 
 25.23 lbR = 
Using now the Law of Sines 
 10 lb 25.23 lb
sin sin110β = ° 
10 lbsin sin110
25.23 lb
β  = °   
 0.3724= 
So: 21.87β = ° 
Angle of inclination of R, φ is then such that: 
30φ β+ = ° 
 8.13φ = ° 
Hence: 25.2 lb=R 8.13°W 
15
 
 
 
 
PROBLEM 2.16 
Solve Problem 2.1 using trigonometry 
Problem 2.1: Two forces are applied to an eye bolt fastened to a beam. 
Determine graphically the magnitude and direction of their resultant 
using (a) the parallelogram law, (b) the triangle rule. 
 
SOLUTION 
Using the force triangle, the Law of Cosines and the Law of Sines 
 
We have: ( )180 50 25α = ° − ° + ° 
 105= ° 
Then: ( ) ( ) ( )( )2 22 4.5 kN 6 kN 2 4.5 kN 6 kN cos105R = + − ° 
 270.226 kN= 
or 8.3801 kNR = 
Now: 8.3801 kN 6 kN
sin105 sin β=° 
6 kNsin sin105
8.3801 kN
β  = °   
 0.6916= 
 43.756β = ° 
 8.38 kN=R 18.76°W 
 
16
 
PROBLEM 2.17 
Solve Problem 2.2 using trigonometry 
Problem 2.2: The cable stays AB and AD help support pole AC. Knowing 
that the tension is 500 N in AB and 160 N in AD, determine graphically 
the magnitude and direction of the resultant of the forces exerted by the 
stays at A using (a) the parallelogram law, (b) the triangle rule. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
From the geometry of the problem: 
1 2tan 38.66
2.5
α −= = ° 
1 1.5tan 30.96
2.5
β −= = ° 
Now: ( )180 38.66 30.96 110.38θ = ° − + ° = 
And, using the Law of Cosines: 
( ) ( ) ( )( )2 22 500 N 160 N 2 500 N 160 N cos110.38R = + − ° 
 2331319 N= 
575.6 NR = 
Using the Law of Sines: 
160 N 575.6 N
sin sin110.38γ = ° 
160 Nsin sin110.38
575.6 N
γ  = °   
 0.2606= 
 15.1γ = ° 
( )90 66.44φ α γ= ° − + = ° 
 576 N=R 66.4°W 
 
17
 
PROBLEM 2.18 
Solve Problem 2.3 using trigonometry 
Problem 2.3: Two forces P and Q are applied as shown at point A of a 
hook support. Knowing that P = 15 lb and Q = 25 lb, determine 
graphically the magnitude and direction of their resultant using (a) the 
parallelogram law, (b) the triangle rule. 
 
SOLUTION 
 
 
 
 
 
Using the force triangle and the Laws of Cosines and Sines 
We have: 
( )180 15 30γ = ° − ° + ° 
 135= ° 
Then: ( ) ( ) ( )( )2 22 15 lb 25 lb 2 15 lb 25 lb cos135R = + − ° 
 21380.3 lb= 
or 37.15 lbR = 
and 
25 lb 37.15 lb
sin sin135β = ° 
25 lbsin sin135
37.15 lb
β  = °   
 0.4758= 
 28.41β = ° 
Then: 75 180α β+ + ° = ° 
76.59α = ° 
 37.2 lb=R 76.6°W 
 
18
 
PROBLEM 2.19 
Two structural members A and B are bolted to a bracket as shown. 
Knowing that both members are in compression and that the force is 
30 kN in member A and 20 kN in member B, determine, using 
trigonometry, the magnitude and direction of the resultant of the forces 
applied to the bracket by members A and B. 
 
 
SOLUTION 
 
 
 
 
 
 
Using the force triangle and the Laws of Cosines and Sines 
We have: ( )180 45 25 110γ = ° − ° + ° = ° 
Then: ( ) ( ) ( )( )2 22 30 kN 20 kN 2 30 kN 20 kN cos110R = + − ° 
 21710.4 kN= 
 41.357 kNR = 
and 
20 kN 41.357 kN
sin sin110α = ° 
20 kNsin sin110
41.357 kN
α  = °   
 0.4544= 
 27.028α = ° 
Hence: 45 72.028φ α= + ° = ° 
 41.4 kN=R 72.0°W 
 
19
 
PROBLEM 2.20 
Two structural members A and B are bolted to a bracket as shown. 
Knowing that both members are in compression and that the force is 
20 kN in member A and 30 kN in member B, determine, using 
trigonometry, the magnitude and direction of the resultant of the forces 
applied to the bracket by members A and B. 
 
SOLUTION 
 
 
 
 
 
Using the force triangle and the Laws of Cosines and Sines 
We have: ( )180 45 25 110γ = ° − ° + ° = ° 
Then: ( ) ( ) ( )( )2 22 30 kN 20 kN 2 30 kN 20 kN cos110R = + − ° 
 21710.4 kN= 
 41.357 kNR = 
and 
30 kN 41.357 kN
sin sin110α = ° 
30 kNsin sin110
41.357 kN
α  = °   
 0.6816= 
 42.97α = ° 
Finally: 45 87.97φ α= + ° = ° 
 41.4 kN=R 88.0°W 
 
 
20
 
 
 
 
PROBLEM 2.21 
Determine the x and y components of each of the forces shown. 
 
SOLUTION 
20 kN Force: 
 ( )20 kN cos 40 ,xF = + ° 15.32 kNxF = W 
 ( )20 kN sin 40 ,yF = + ° 12.86 kNyF = W 
30 kN Force: 
 ( )30 kN cos70 ,xF = − ° 10.26 kNxF = − W 
 ( )30 kN sin 70 ,yF = + ° 28.2 kNyF = W 
42 kN Force: 
 ( )42 kN cos 20 ,xF = − ° 39.5 kNxF = − W 
 ( )42 kN sin 20 ,yF = + ° 14.36 kNyF = W 
21
 
 
 
 
PROBLEM 2.22 
Determine the x and y components of each of the forces shown. 
 
SOLUTION 
40 lb Force: 
 ( )40 lb sin 50 ,xF = − ° 30.6 lbxF = − W 
 ( )40 lb cos50 ,yF = − ° 25.7 lbyF = − W 
60 lb Force: 
 ( )60 lb cos60 ,xF = + ° 30.0 lbxF = W 
 ( )60 lb sin 60 ,yF = − ° 52.0 lbyF = − W 
80 lb Force: 
 ( )80 lb cos 25 ,xF = + ° 72.5 lbxF = W 
 ( )80 lb sin 25 ,yF = + ° 33.8 lbyF = W 
 
22
 
 
 
 
PROBLEM 2.23 
Determine the x and y components of each of the forces shown. 
 
SOLUTION 
 
 
 
 
 
We compute the following distances: 
( ) ( )
( ) ( )
( ) ( )
2 2
2 2
2 2
48 90 102 in.
56 90 106 in.
80 60 100 in.
OA
OB
OC
= + =
= + =
= + =
 
Then: 
204 lb Force: 
 ( ) 48102 lb ,
102x
F = − 48.0 lbxF = − W 
 ( ) 90102 lb ,
102y
F = + 90.0 lbyF = W 
212 lb Force: 
 ( ) 56212 lb ,
106x
F = + 112.0 lbxF = W 
 ( ) 90212 lb ,
106y
F = + 180.0 lbyF = W 
400 lb Force: 
 ( ) 80400 lb ,
100x
F = − 320 lbxF = − W 
 ( ) 60400 lb ,
100y
F = − 240 lbyF = − W 
23
 
 
 
PROBLEM 2.24 
Determine the x and y components of each of the forces shown. 
 
SOLUTION 
 
 
 
 
 
We compute the following distances: 
 ( ) ( )2 270 240 250 mmOA = + = 
( ) ( )2 2210 200 290 mmOB = + = 
 ( ) ( )2 2120 225 255 mmOC = + = 
500 N Force: 
 70500 N
250x
F  = −    140.0 NxF = − W 
 240500 N
250y
F  = +    480 NyF = W 
435 N Force: 
 210435 N
290x
F  = +    315 NxF = W 
 200435 N
290y
F  = +    300 NyF = W 
510 N Force: 
 120510 N
255x
F  = +    240 NxF = W 
 225510 N
255y
F  = −    450 NyF = − W 
24
 
 
 
 
 
PROBLEM 2.25 
While emptying a wheelbarrow, a gardener exerts on each handle AB a 
force P directed along line CD. Knowing that P must have a 135-N 
horizontal component, determine (a) the magnitude of the force P, (b) its 
vertical component. 
 
SOLUTION 
 
(a) 
cos 40
xPP = ° 
 135 N
cos 40
= ° 
 or 176.2 NP = W 
(b) tan 40 sin 40y xP P P= ° = ° 
 ( )135 N tan 40= °
or 113.3 NyP = W 
25
 
 
 
 
PROBLEM 2.26 
Member BD exerts on member ABC a force P directed along line BD. 
Knowing that P must have a 960-N vertical component, determine (a) the 
magnitude of the force P, (b) its horizontal component. 
 
SOLUTION 
 
(a) 
sin 35
yPP = ° 
 960 N
sin 35
= ° 
 or 1674 NP = W 
(b) 
tan 35
y
x
P
P = ° 
 960 N
tan 35
= ° 
 or 1371 NxP = W 
26
 
 
 
 
 
PROBLEM 2.27 
Member CB of the vise shown exerts on block B a force P directed along 
line CB. Knowing that P must have a 260-lb horizontal component, 
determine (a) the magnitude of the force P, (b) its vertical component. 
 
SOLUTION 
 
We note: 
CB exerts force P on B along CB, and the horizontal component of P is 260 lb.xP = 
Then: 
(a) sin 50xP P= ° 
 
sin 50
xPP = ° 
 260 lb
sin50
= ° 
 339.4 lb= 339 lbP = W 
(b) tan 50x yP P= ° 
 
tan 50
x
y
PP = ° 
 260 lb
tan 50
= ° 
 218.2 lb= 218 lby =P W 
27
 
 
 
 
PROBLEM 2.28 
Activator rod AB exerts on crank BCD a force P directed along line AB. 
Knowing that P must have a 25-lb component perpendicular to arm BC of 
the crank, determine (a) the magnitude of the force P, (b) its component 
along line BC. 
 
SOLUTION 
 
Using the x and y axes shown. 
(a) 25 lbyP = 
 Then: 
sin 75
yPP = ° 
 25 lb
sin 75
= ° 
 or 25.9 lbP = W 
(b) 
tan 75
y
x
P
P = ° 
 25 lb
tan 75
= ° 
 or 6.70 lbxP = W 
28
 
 
 
 
 
PROBLEM 2.29 
The guy wire BD exerts on the telephone pole AC a force P directed 
along BD. Knowing that P has a 450-N component along line AC, 
determine (a) the magnitude of the force P, (b) its component in a 
direction perpendicular to AC. 
 
SOLUTION 
 
Note that the force exerted by BD on the pole is directed along BD, and the component of P along AC 
is 450 N. 
Then: 
(a) 450 N 549.3 N
cos35
P = =° 
 549 NP = W 
(b) ( )450 N tan 35xP = ° 
 315.1 N= 
 315 NxP = W 
29
 
 
 
 
PROBLEM 2.30 
The guy wire BD exerts on the telephone pole AC a force P directed 
along BD. Knowing that P has a 200-N perpendicular to the pole AC, 
determine (a) the magnitude of the force P, (b) its component along 
line AC. 
 
SOLUTION 
 
(a) 
sin 38
xPP = ° 
 200 N
sin 38
= ° 
 324.8 N= or 325 NP = W 
(b) 
tan 38
x
y
PP = ° 
 200 N
tan 38
= ° 
 255.98 N= 
 or 256 NyP = W 
 
30
 
 
 
 
 
PROBLEM 2.31 
Determine the resultant of the three forces of Problem 2.24. 
Problem 2.24: Determine the x and y components of each of the forces 
shown. 
 
SOLUTION 
 
From Problem 2.24: 
( ) ( )500 140 N 480 N= − +F i j 
( ) ( )425 315 N 300 N= +F i j 
( ) ( )510 240 N 450 N= −F i j 
( ) ( )415 N 330 N= Σ = +R F i j 
Then: 
1 330tan 38.5
415
α −= = ° 
( ) ( )2 2415 N 330 N 530.2 NR = + = 
Thus: 530 N=R 38.5°W 
31
 
 
 
 
PROBLEM 2.32 
Determine the resultant of the three forces of Problem 2.21. 
Problem 2.21: Determine the x and y components of each of the forces 
shown. 
 
SOLUTION 
 
From Problem 2.21: 
( ) ( )20 15.32 kN 12.86 kN= +F i j 
( ) ( )30 10.26 kN 28.2 kN= − +F i j 
( ) ( )42 39.5 kN 14.36 kN= − +F i j 
( ) ( )34.44 kN 55.42 kN= Σ = − +R F i j 
Then: 
1 55.42tan 58.1
34.44
α −= = °− 
( ) ( )2 255.42 kN 34.44 N 65.2 kNR = + − = 
 65.2 kNR = 58.2°W 
32
 
 
 
 
 
PROBLEM 2.33 
Determine the resultant of the three forces of Problem 2.22. 
Problem 2.22: Determine the x and y components of each of the forces 
shown. 
 
SOLUTION 
The components of the forces were determined in 2.23. 
 
 
 
 
 
 
 
 x yR R= +R i j 
 ( ) ( )71.9 lb 43.86 lb= −i j 
43.86tan
71.9
α = 
 31.38α = ° 
( ) ( )2 271.9 lb 43.86 lbR = + − 
 84.23 lb= 
 84.2 lb=R 31.4°W 
Force comp. (lb)x comp. (lb)y 
40 lb 30.6− 25.7− 
60 lb 30 51.96− 
80 lb 72.5 33.8 
 71.9xR = 43.86yR = − 
33
 
 
 
 
PROBLEM 2.34 
Determine the resultant of the three forces of Problem 2.23. 
Problem 2.23: Determine the x and y components of each of the forces 
shown. 
 
SOLUTION 
The components of the forces were 
determined in Problem 2.23. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
( ) ( )204 48.0 lb 90.0 lb= − +F i j 
( ) ( )212 112.0 lb 180.0 lb= +F i j 
( ) ( )400 320 lb 240 lb= − −F i j 
Thus 
x y= +R R R 
( ) ( )256 lb 30.0 lb= − +R i j 
Now: 
30.0tan
256
α = 
1 30.0tan 6.68
256
α −= = ° 
and 
( ) ( )2 2256 lb 30.0 lbR = − + 
 257.75 lb= 
 258 lb=R 6.68°W 
34
 
 
 
 
 
PROBLEM 2.35 
Knowing that 35 ,α = ° determine the resultant of the three forces 
shown. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
300-N Force: 
( )300 N cos 20 281.9 NxF = ° = 
( )300 N sin 20 102.6 NyF = ° = 
400-N Force: 
( )400 N cos55 229.4 NxF = ° = 
( )400 N sin 55 327.7 NyF = ° = 
600-N Force: 
( )600 N cos35 491.5 NxF = ° = 
( )600 N sin 35 344.1 NyF = − ° = − 
and 
1002.8 Nx xR F= Σ = 
86.2 Ny yR F= Σ = 
( ) ( )2 21002.8 N 86.2 N 1006.5 NR = + = 
Further: 
86.2tan
1002.8
α = 
1 86.2tan 4.91
1002.8
α −= = ° 
 1007 N=R 4.91°W 
35
 
 
 
 
PROBLEM 2.36 
Knowing that 65 ,α = ° determine the resultant of the three forces 
shown. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
300-N Force: 
( )300 N cos 20 281.9 NxF = ° = 
( )300 N sin 20 102.6 NyF = ° = 
400-N Force: 
( )400 N cos85 34.9 NxF = ° = 
( )400 N sin85 398.5 NyF = ° = 
600-N Force: 
( )600 N cos5 597.7 NxF = ° = 
( )600 N sin 5 52.3 NyF = − ° = − 
and 
914.5 Nx xR F= Σ = 
448.8 Ny yR F= Σ = 
( ) ( )2 2914.5 N 448.8 N 1018.7 NR = + = 
Further: 
448.8tan
914.5
α = 
1 448.8tan 26.1
914.5
α −= = ° 
 1019 N=R 26.1°W 
 
36
 
 
PROBLEM 2.37 
Knowing that the tension in cable BC is 145 lb, determine the resultant of 
the three forces exerted at point B of beam AB. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
 
 
 
 
Cable BC Force: 
( ) 84145 lb 105 lb
116x
F = − = − 
( ) 80145 lb 100 lb
116y
F = = 
100-lb Force: 
( ) 3100 lb 60 lb
5x
F = − = − 
( ) 4100 lb 80 lb
5y
F = − = − 
156-lb Force: 
( )12156 lb 144 lb
13x
F = = 
( ) 5156 lb 60 lb
13y
F = − = − 
and 
21 lb, 40 lbx x y yR F R F= Σ = − = Σ = − 
( ) ( )2 221 lb 40 lb 45.177 lbR = − + − = 
Further: 
40tan
21
α = 
1 40tan 62.3
21
α −= = ° 
Thus: 45.2 lb=R 62.3°W 
 
37
 
PROBLEM 2.38 
Knowing that 50 ,α = ° determine the resultant of the three forces 
shown. 
 
SOLUTION 
 
 
 
 
 
 
The resultant force R has the x- and y-components: 
( ) ( ) ( )140 lb cos50 60 lb cos85 160 lb cos50x xR F= Σ = ° + ° − ° 
 7.6264 lbxR = − 
and 
( ) ( ) ( )140 lb sin 50 60 lb sin85 160 lb sin 50y yR F= Σ = ° + ° + ° 
 289.59 lbyR = 
Further: 
290tan
7.6
α = 
1 290tan 88.5
7.6
α −= = ° 
Thus: 290 lb=R 88.5°W 
 
 
 
 
 
 
 
 
 
 
 
 
 
38
 
 
PROBLEM 2.39 
Determine (a) the required value of α if the resultant of the three forces 
shown is to be vertical, (b) the corresponding magnitude of the resultant. 
 
SOLUTION 
For an arbitrary angle ,α we have: 
( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cosx xR F α α α= Σ = + + ° − 
(a) So, for R to be vertical: 
( ) ( ) ( ) ( )140 lb cos 60 lb cos 35 160 lb cos 0x xR F α α α= Σ = + + ° − = 
 Expanding, 
( )cos 3 cos cos35
sin sin 35 0α α α− + ° − ° = 
 Then: 
1
3cos35tan
sin 35
α ° −= ° 
 or 
 
1
1 3cos35tan 40.265
sin 35
α −  ° −= = °  ° 
 40.3α = °W 
(b) Now: 
( ) ( ) ( )140 lb sin 40.265 60 lb sin 75.265 160 lb sin 40.265y yR R F= = Σ = ° + ° + ° 
 252 lbR R= = W 
 
39
 
PROBLEM 2.40 
For the beam of Problem 2.37, determine (a) the required tension in cable 
BC if the resultant of the three forces exerted at point B is to be vertical, 
(b) the corresponding magnitude of the resultant. 
Problem 2.37: Knowing that the tension in cable BC is 145 lb, determine 
the resultant of the three forces exerted at point B of beam AB. 
 
 
SOLUTION 
We have: 
( ) ( )84 12 3156 lb 100 lb
116 13 5x x BC
R F T= Σ = − + − 
or 0.724 84 lbx BCR T= − + 
and 
( ) ( )80 5 4156 lb 100 lb
116 13 5y y BC
R F T= Σ = − − 
 0.6897 140 lby BCR T= − 
(a) So, for R to be vertical, 
0.724 84 lb 0x BCR T= − + = 
 116.0 lbBCT = W 
(b) Using 
116.0 lbBCT = 
( )0.6897 116.0 lb 140 lb 60 lbyR R= = − = − 
 60.0 lbR R= = W 
 
 
40
 
 
 
 
 
PROBLEM 2.41 
Boom AB is held in the position shown by three cables. Knowing that the 
tensions in cables AC and AD are 4 kN and 5.2 kN, respectively, 
determine (a) the tension in cable AE if the resultant of the tensions 
exerted at point A of the boom must be directed along AB, 
(b) the corresponding magnitude of the resultant. 
 
SOLUTION 
 
 
 
 
 
 
 
 
Choose x-axis along bar AB. 
Then 
(a) Require 
( ) ( )0: 4 kN cos 25 5.2 kN sin 35 sin 65 0y y AER F T= Σ = ° + ° − ° = 
 or 7.2909 kNAET = 
 7.29 kNAET = W 
(b) xR F= Σ 
 ( ) ( ) ( )4 kN sin 25 5.2 kN cos35 7.2909 kN cos65= − ° − ° − ° 
 9.03 kN= − 
 9.03 kNR = W 
41
 
 
 
 
PROBLEM 2.42 
For the block of Problems 2.35 and 2.36, determine (a) the required value 
of α of the resultant of the three forces shown is to be parallel to the 
incline, (b) the corresponding magnitude of the resultant. 
Problem 2.35: Knowing that 35 ,α = ° determine the resultant of the 
three forces shown. 
Problem 2.36: Knowing that 65 ,α = ° determine the resultant of the 
three forces shown. 
 
SOLUTION 
 
 
 
 
 
 
 
Selecting the x axis along ,aa′ we write 
 ( ) ( )300 N 400 N cos 600 N sinx xR F α α= Σ = + + (1) 
 ( ) ( )400 N sin 600 N cosy yR F α α= Σ = − (2) 
(a) Setting 0yR = in Equation (2): 
 Thus 600tan 1.5
400
α = = 
 56.3α = °W 
(b) Substituting for α in Equation (1): 
( ) ( )300 N 400 N cos56.3 600 N sin 56.3xR = + ° + ° 
 1021.1 NxR = 
 1021 NxR R= = W 
42
 
 
 
PROBLEM 2.43 
Two cables are tied together at C and are loaded as shown. Determine the 
tension (a) in cable AC, (b) in cable BC. 
 
SOLUTION 
Free-Body Diagram 
 
 
 
 
 
 
 
From the geometry, we calculate the distances: 
( ) ( )2 216 in. 12 in. 20 in.AC = + = 
( ) ( )2 220 in. 21 in. 29 in.BC = + = 
Then, from the Free Body Diagram of point C: 
 16 210: 0
20 29x AC BC
F T TΣ = − + = 
or 29 4
21 5BC AC
T T= × 
and 12 200: 600 lb 0
20 29y AC BC
F T TΣ = + − = 
or 12 20 29 4 600 lb 0
20 29 21 5AC AC
T T + × − =   
Hence: 440.56 lbACT = 
(a) 441 lbACT = W 
(b) 487 lbBCT = W 
43
 
 
 
 
PROBLEM 2.44 
Knowing that 25 ,α = ° determine the tension (a) in cable AC, (b) in 
rope BC. 
 
SOLUTION 
Free-Body Diagram Force Triangle 
 
Law of Sines: 
5 kN
sin115 sin 5 sin 60
AC BCT T= =° ° ° 
(a) 5 kN sin115 5.23 kN
sin 60AC
T = ° =° 5.23 kNACT = W 
(b) 5 kN sin 5 0.503 kN
sin 60BC
T = ° =° 0.503 kNBCT = W 
44
 
 
 
 
 
PROBLEM 2.45 
Knowing that 50α = ° and that boom AC exerts on pin C a force 
directed long line AC, determine (a) the magnitude of that force, (b) the 
tension in cable BC. 
 
SOLUTION 
Free-Body Diagram Force Triangle 
 
Law of Sines: 
 400 lb
sin 25 sin 60 sin 95
AC BCF T= =° ° ° 
(a) 400 lb sin 25 169.69 lb
sin 95AC
F = ° =° 169.7 lbACF = W 
(b) 400 sin 60 347.73 lb
sin 95BC
T = ° =° 348 lbBCT = W 
45
 
 
 
 
PROBLEM 2.46 
Two cables are tied together at C and are loaded as shown. Knowing that 
30 ,α = ° determine the tension (a) in cable AC, (b) in cable BC. 
 
SOLUTION 
Free-Body Diagram Force Triangle 
 
Law of Sines: 
2943 N
sin 60 sin 55 sin 65
AC BCT T= =° ° ° 
(a) 2943 N sin 60 2812.19 N
sin 65AC
T = ° =° 2.81 kNACT = W 
(b) 2943 N sin 55 2659.98 N
sin 65BC
T = ° =° 2.66 kNBCT = W 
46
 
 
 
 
 
PROBLEM 2.47 
A chairlift has been stopped in the position shown. Knowing that each 
chair weighs 300 N and that the skier in chair E weighs 890 N, determine 
that weight of the skier in chair F. 
 
SOLUTION 
 Free-Body Diagram Point B 
 
 
 Force Triangle 
 
 Free-Body Diagram Point C 
 
 Force Triangle 
 
 
In the free-body diagram of point B, the geometry gives: 
1 9.9tan 30.51
16.8AB
θ −= = ° 
1 12tan 22.61
28.8BC
θ −= = ° 
 
 
Thus, in the force triangle, by the Law of Sines: 
1190 N
sin 59.49 sin 7.87
BCT =° ° 
7468.6 NBCT = 
 
In the free-body diagram of point C (with W the sum of weights of chair 
and skier) the geometry gives: 
1 1.32tan 10.39
7.2CD
θ −= = ° 
Hence, in the force triangle, by the Law of Sines: 
7468.6 N
sin12.23 sin100.39
W =° ° 
1608.5 NW = 
Finally, the skier weight 1608.5 N 300 N 1308.5 N= − = 
 skier weight 1309 N= W 
47
 
 
 
 
PROBLEM 2.48 
A chairlift has been stopped in the position shown. Knowing that each 
chair weighs 300 N and that the skier in chair F weighs 800 N, determine 
the weight of the skier in chair E. 
 
SOLUTION 
 Free-Body Diagram Point F 
 
 Force Triangle 
 
 
 Free-Body Diagram Point E 
 
 Force Triangle 
 
 
In the free-body diagram of point F, the geometry gives: 
1 12tan 22.62
28.8EF
θ −= = ° 
1 1.32tan 10.39
7.2DF
θ −= = ° 
Thus, in the force triangle, by the Law of Sines: 
1100 N
sin100.39 sin12.23
EFT =° ° 
5107.5 NBCT = 
In the free-body diagram of point E (with W the sum of weights of chair 
and skier) the geometry gives: 
1 9.9tan 30.51
16.8AE
θ −= = ° 
Hence, in the force triangle, by the Law of Sines: 
5107.5 N
sin 7.89 sin 59.49
W =° ° 
813.8 NW = 
Finally, the skier weight 813.8 N 300 N 513.8 N= − = 
 skier weight 514 N= W 
48
 
 
 
 
 
PROBLEM 2.49 
Four wooden members are joined with metal plate connectors and are in 
equilibrium under the action of the four fences shown. Knowing that 
FA = 510 lb and FB = 480 lb, determine the magnitudes of the other two 
forces. 
 
SOLUTION 
Free-Body Diagram 
 
 
 
 
 
 
 
Resolving the forces into x and y components: 
( ) ( )0: 510 lb sin15 480 lb cos15 0x CF FΣ = + ° − ° = 
 or 332 lbCF = W 
( ) ( )0: 510 lb cos15 480 lb sin15 0y DF FΣ = − ° + ° = 
 or 368 lbDF = W 
49
 
 
 
 
PROBLEM 2.50 
Four wooden members are joined with metal plate connectors and are in 
equilibrium under the action of the four fences shown. Knowing that 
FA = 420 lb and FC = 540 lb, determine the magnitudes of the other two 
forces. 
 
SOLUTION 
 
 
 
 
 
 
 
 
 
Resolving the forces into x and y components: 
( ) ( )0: cos15 540 lb 420 lb cos15 0 or 671.6 lbx B BF F FΣ = − ° + + ° = = 
 672 lbBF = W 
( ) ( )0: 420 lb cos15 671.6 lb sin15 0y DF FΣ = − ° + ° = 
 or 232 lbDF = W 
 
50
 
 
 
 
PROBLEM 2.51 
Two forces P and Q are applied as shown to an aircraft connection. 
Knowing that
the connection is in equilibrium and the P = 400 lb and 
Q = 520 lb, determine the magnitudes of the forces exerted on the rods 
A and B. 
 
SOLUTION 
Free-Body Diagram 
 
 
 
 
Resolving the forces into x and y directions: 
0A B= + + + =R P Q F F 
Substituting components: 
( ) ( ) ( )400 lb 520 lb cos55 520 lb sin 55   = − + ° − °   R j i j 
 ( ) ( )cos55 sin 55 0B A AF F F+ − ° + ° =i i j 
In the y-direction (one unknown force) 
( )400 lb 520 lb sin 55 sin 55 0AF− − ° + ° = 
Thus, 
( )400 lb 520 lb sin 55 1008.3 lb
sin 55A
F
+ °= =° 
 1008 lbAF = W 
In the x-direction: 
( )520 lb cos55 cos55 0B AF F° + − ° = 
Thus, 
 ( )cos55 520 lb cos55B AF F= ° − ° 
 ( ) ( )1008.3 lb cos55 520 lb cos55= ° − ° 
 280.08 lb= 
 280 lbBF = W 
51
 
 
 
 
PROBLEM 2.52 
Two forces P and Q are applied as shown to an aircraft connection. 
Knowing that the connection is in equilibrium and that the magnitudes of 
the forces exerted on rods A and B are FA = 600 lb and FB = 320 lb, 
determine the magnitudes of P and Q. 
 
SOLUTION 
Free-Body Diagram 
 
 
 
 
Resolving the forces into x and y directions: 
0A B= + + + =R P Q F F 
Substituting components: 
( ) ( ) ( )320 lb 600 lb cos55 600 lb sin 55   = − ° + °   R i i j 
 ( ) ( )cos55 sin 55 0P Q Q+ + ° − ° =i i j 
In the x-direction (one unknown force) 
( )320 lb 600 lb cos55 cos55 0Q− ° + ° = 
Thus, 
( )320 lb 600 lb cos55 42.09 lb
cos55
Q
− + °= =° 
 42.1 lbQ = W 
In the y-direction: 
( )600 lb sin 55 sin 55 0P Q° − − ° = 
Thus, 
( )600 lb sin 55 sin 55 457.01 lbP Q= ° − ° = 
 457 lbP = W 
52
 
 
 
 
PROBLEM 2.53 
Two cables tied together at C are loaded as shown. Knowing that 
W = 840 N, determine the tension (a) in cable AC, (b) in cable BC. 
 
SOLUTION 
Free-Body Diagram 
 
 
 
 
From geometry: 
The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. 
The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. 
Thus: 
 ( )3 15 150: 680 N 0
5 17 17x CA CB
F T TΣ = − + − = 
or 
 1 5 200 N
5 17CA CB
T T− + = (1) 
and 
 ( )4 8 80: 680 N 840 N 05 17 17y CA CBF T TΣ = + − − = 
or 
 1 2 290 N
5 17CA CB
T T+ = (2) 
Solving Equations (1) and (2) simultaneously: 
(a) 750 NCAT = W 
(b) 1190 NCBT = W 
53
 
 
 
 
PROBLEM 2.54 
Two cables tied together at C are loaded as shown. Determine the range 
of values of W for which the tension will not exceed 1050 N in either 
cable. 
 
SOLUTION 
 Free-Body Diagram 
 
 
 
 
From geometry: 
The sides of the triangle with hypotenuse CB are in the ratio 8:15:17. 
The sides of the triangle with hypotenuse CA are in the ratio 3:4:5. 
Thus: 
 ( )3 15 150: 680 N 0
5 17 17x CA CB
F T TΣ = − + − = 
or 
 1 5 200 N
5 17CA CB
T T− + = (1) 
and 
 ( )4 8 80: 680 N 05 17 17y CA CBF T T WΣ = + − − = 
or 
 1 2 180 N
5 17 4
+ = +CA CBT T W (2) 
Then, from Equations (1) and (2) 
17680 N
28
25
28
CB
CA
T W
T W
= +
=
 
Now, with 1050 NT ≤ 
25: 1050 N
28CA CA
T T W= = 
or 1176 NW = 
and 
17: 1050 N 680 N
28CB CB
T T W= = + 
or 609 NW = 0 609 N∴ ≤ ≤W W 
54
 
 
 
 
PROBLEM 2.55 
The cabin of an aerial tramway is suspended from a set of wheels that can 
roll freely on the support cable ACB and is being pulled at a constant 
speed by cable DE. Knowing that 40α = ° and β = 35°, that the 
combined weight of the cabin, its support system, and its passengers is 
24.8 kN, and assuming the tension in cable DF to be negligible, 
determine the tension (a) in the support cable ACB, (b) in the traction 
cable DE. 
 
SOLUTION 
 
 
 
 
 
Note: In Problems 2.55 and 2.56 the cabin is considered as a particle. If 
considered as a rigid body (Chapter 4) it would be found that its center of 
gravity should be located to the left of the centerline for the line CD to be 
vertical. 
Now 
 ( )0: cos35 cos 40 cos 40 0x ACB DEF T TΣ = ° − ° − ° = 
or 
 0.0531 0.766 0ACB DET T− = (1) 
and 
 ( )0: sin 40 sin 35 sin 40 24.8 kN 0y ACB DEF T TΣ = ° − ° + ° − = 
or 
 0.0692 0.643 24.8 kNACB DET T+ = (2) 
From (1) 
14.426ACB DET T= 
Then, from (2) 
( )0.0692 14.426 0.643 24.8 kNDE DET T+ = 
and 
 (b) 15.1 kNDET = W 
 (a) 218 kNACBT = W 
55
 
 
 
PROBLEM 2.56 
The cabin of an aerial tramway is suspended from a set of wheels that can 
roll freely on the support cable ACB and is being pulled at a constant 
speed by cable DE. Knowing that 42α = ° and β = 32°, that the tension 
in cable DE is 20 kN, and assuming the tension in cable DF to be 
negligible, determine (a) the combined weight of the cabin, its support 
system, and its passengers, (b) the tension in the support cable ACB. 
 
SOLUTION 
Free-Body Diagram 
 
First, consider the sum of forces in the x-direction because there is only one unknown force: 
 ( ) ( )0: cos32 cos 42 20 kN cos 42 0x ACBF TΣ = ° − ° − ° = 
or 
0.1049 14.863 kNACBT = 
 (b) 141.7 kNACBT = W 
Now 
 ( ) ( )0: sin 42 sin 32 20 kN sin 42 0y ACBF T WΣ = ° − ° + ° − = 
or 
 ( )( ) ( )( )141.7 kN 0.1392 20 kN 0.6691 0W+ − = 
 (a) 33.1 kNW = W 
56
 
 
 
PROBLEM 2.57 
A block of weight W is suspended from a 500-mm long cord and two 
springs of which the unstretched lengths are 450 mm. Knowing that the 
constants of the springs are kAB = 1500 N/m and kAD = 500 N/m, 
determine (a) the tension in the cord, (b) the weight of the block. 
 
SOLUTION 
Free-Body Diagram At A 
 
 
 
 
First note from geometry: 
The sides of the triangle with hypotenuse AD are in the ratio 8:15:17. 
The sides of the triangle with hypotenuse AB are in the ratio 3:4:5. 
The sides of the triangle with hypotenuse AC are in the ratio 7:24:25. 
Then: 
( )AB AB AB oF k L L= − 
and 
( ) ( )2 20.44 m 0.33 m 0.55 mABL = + = 
So: 
 ( )1500 N/m 0.55 m 0.45 mABF = − 
 150 N= 
Similarly, 
( )AD AD AD oF k L L= − 
Then: 
 ( ) ( )2 20.66 m 0.32 m 0.68 mADL = + = 
 ( )1500 N/m 0.68 m 0.45 mADF = − 
 115 N= 
(a) 
 ( ) ( )4 7 150: 150 N 115 N 0
5 25 17x AC
F TΣ = − + − = 
 or 
 66.18 NACT = 66.2 NACT = W 
57
 
 
 
PROBLEM 2.57 CONTINUED 
(b) and 
 ( ) ( ) ( )3 24 80: 150 N 66.18 N 115 N 05 25 17yF WΣ = + + − = 
 or 208 N=W W 
 
58
 
 
 
 
PROBLEM 2.58 
A load of weight 400 N is suspended from a spring and two cords which 
are attached to blocks of weights 3W and W as shown. Knowing that the 
constant of the spring is 800 N/m, determine (a) the value of W, (b) the 
unstretched length of the spring. 
 
SOLUTION 
Free-Body Diagram At A 
 
 
 
 
First note from geometry: 
The sides of the triangle with hypotenuse AD are in the ratio 12:35:37. 
The sides of the triangle with hypotenuse AC are in the ratio 3:4:5. 
The sides of the triangle with hypotenuse AB are also in the ratio 
12:35:37. 
Then: 
 ( ) ( )4 35 120: 3 0
5 37 37x s
F W W FΣ = − + + = 
or 
4.4833sF W= 
and 
 ( ) ( )3 12 350: 3 400 N 05 37 37y sF W W FΣ = + + − = 
Then: 
( ) ( ) ( )3 12 353 4.4833 400 N 0
5 37 37
W W W+ + − = 
or 
 62.841 NW = 
and 
281.74 NsF = 
or 
(a) 62.8 NW = W 
 
59
 
 
 
PROBLEM 2.58 CONTINUED 
(b) Have spring force 
( )s AB oF k L L= − 
 Where 
( )AB AB AB oF k L L= − 
 and 
( ) ( )2 20.360 m 1.050 m 1.110 mABL = + = 
 So: 
( )0281.74 N 800 N/m 1.110 mL= − 
 or 0 758 mmL = W 
 
60
 
 
 
 
 
PROBLEM 2.59 
For the cables and loading of Problem 2.46, determine (a) the value of α 
for which the tension in cable BC is as small as possible, (b) the 
corresponding value of
the tension. 
 
SOLUTION 
The smallest BCT is when BCT is perpendicular to the direction of ACT 
Free-Body Diagram At C Force Triangle 
 
(a) 55.0α = °W 
(b) ( )2943 N sin 55BCT = ° 
 2410.8 N= 
 2.41 kNBCT = W 
61
 
 
 
 
PROBLEM 2.60 
Knowing that portions AC and BC of cable ACB must be equal, determine 
the shortest length of cable which can be used to support the load shown 
if the tension in the cable is not to exceed 725 N. 
 
SOLUTION 
Free-Body Diagram: C 
( )For 725 NT = 
 
 
 
 
 
 
 
 0: 2 1000 N 0y yF TΣ = − = 
500 NyT = 
2 2 2
x yT T T+ = 
( ) ( )2 22 500 N 725 NxT + = 
525 NxT = 
 
By similar triangles: 
1.5 m
725 525
BC = 
2.07 m∴ =BC 
( )2 4.14 mL BC= = 
 4.14 mL = W 
 
62
 
 
 
 
 
PROBLEM 2.61 
Two cables tied together at C are loaded as shown. Knowing that the 
maximum allowable tension in each cable is 200 lb, determine (a) the 
magnitude of the largest force P which may be applied at C, (b) the 
corresponding value of α. 
 
SOLUTION 
 Free-Body Diagram: C Force Triangle 
 
 
 
 
 
 
Force triangle is isoceles with 
2 180 85β = ° − ° 
47.5β = ° 
(a) ( )2 200 lb cos 47.5 270 lbP = ° = 
 Since 0,P > the solution is correct. 270 lbP = W 
(b) 180 55 47.5 77.5α = ° − ° − ° = ° 77.5α = °W 
63
 
 
 
 
PROBLEM 2.62 
Two cables tied together at C are loaded as shown. Knowing that the 
maximum allowable tension is 300 lb in cable AC and 150 lb in cable BC, 
determine (a) the magnitude of the largest force P which may be applied 
at C, (b) the corresponding value of α. 
 
SOLUTION 
 Free-Body Diagram: C Force Triangle 
 
 
 
 
 
 
 
(a) Law of Cosines: 
( ) ( ) ( )( )2 22 300 lb 150 lb 2 300 lb 150 lb cos85P = + − ° 
 323.5 lbP = 
 Since 300 lb,P > our solution is correct. 324 lbP = W 
(b) Law of Sines: 
sin sin85
300 323.5
β °= ° 
sin 0.9238β = 
 or 67.49β = ° 
180 55 67.49 57.5α = ° − ° − ° = ° 
 57.5α = °W 
64
 
 
 
 
 
PROBLEM 2.63 
For the structure and loading of Problem 2.45, determine (a) the value of 
α for which the tension in cable BC is as small as possible, (b) the 
corresponding value of the tension. 
 
SOLUTION 
BCT must be perpendicular to ACF to be as small as possible. 
 Free-Body Diagram: C Force Triangle is 
 a right triangle 
 
 
 
 
 
 
 
(a) We observe: 55α = ° 55α = °W 
(b) ( )400 lb sin 60BCT = ° 
 or 346.4 lbBCT = 346 lbBCT = W 
 
65
 
 
 
 
PROBLEM 2.64 
Boom AB is supported by cable BC and a hinge at A. Knowing that the 
boom exerts on pin B a force directed along the boom and that the tension 
in rope BD is 70 lb, determine (a) the value of α for which the tension in 
cable BC is as small as possible, (b) the corresponding value of the 
tension. 
 
SOLUTION 
Free-Body Diagram: B 
 
 
 
(a) Have: 0BD AB BC+ + =T F T 
 where magnitude and direction of BDT are known, and the direction 
 of ABF is known. 
 
 Then, in a force triangle: 
 By observation, BCT is minimum when 90.0α = °W 
(b) Have ( ) ( )70 lb sin 180 70 30BCT = ° − ° − ° 
 68.93 lb= 
 68.9 lbBCT = W 
66
 
 
 
 
 
PROBLEM 2.65 
Collar A shown in Figure P2.65 and P2.66 can slide on a frictionless 
vertical rod and is attached as shown to a spring. The constant of the 
spring is 660 N/m, and the spring is unstretched when h = 300 mm. 
Knowing that the system is in equilibrium when h = 400 mm, determine 
the weight of the collar. 
 
SOLUTION 
Free-Body Diagram: Collar A 
 
 
 
Have: ( )s AB ABF k L L′= − 
where: 
( ) ( )2 20.3 m 0.4 m 0.3 2 mAB ABL L′ = + = 
 0.5 m= 
Then: ( )660 N/m 0.5 0.3 2 msF = − 
 49.986 N= 
For the collar: 
 ( )40: 49.986 N 0
5y
F WΣ = − + = 
 or 40.0 NW = W 
67
 
 
 
 
PROBLEM 2.66 
The 40-N collar A can slide on a frictionless vertical rod and is attached 
as shown to a spring. The spring is unstretched when h = 300 mm. 
Knowing that the constant of the spring is 560 N/m, determine the value 
of h for which the system is in equilibrium. 
 
SOLUTION 
Free-Body Diagram: Collar A 
 
 
 ( )2 20: 00.3y s
hF W F
h
Σ = − + =
+
 
or 240 0.09shF h= + 
Now.. ( )s AB ABF k L L′= − 
where ( )2 20.3 m 0.3 2 mAB ABL h L′ = + = 
Then: ( )2 2560 0.09 0.3 2 40 0.09h h h + − = +   
or ( ) 214 1 0.09 4.2 2 mh h h h− + = ∼ 
Solving numerically, 
 415 mmh = W 
 
68
 
 
PROBLEM 2.67 
A 280-kg crate is supported by several rope-and-pulley arrangements as 
shown. Determine for each arrangement the tension in the rope. (Hint: 
The tension in the rope is the same on each side of a simple pulley. This 
can be proved by the methods of Chapter 4.) 
 
SOLUTION 
Free-Body Diagram of pulley 
(a) 
 
 
 
 
(b) 
 
 
 
 
(c) 
 
 
 
 
(d) 
 
 
 
 
 
 
(e) 
 
 
 
 
 
 
 ( )( )20: 2 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
2
T = 
 1373 NT = W 
 ( )( )20: 2 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
2
T = 
 1373 NT = W 
 ( )( )20: 3 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
3
T = 
 916 NT = W 
 ( )( )20: 3 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
3
T = 
 916 NT = W 
 
 ( )( )20: 4 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
4
T = 
 687 NT = W 
 
69
 
PROBLEM 2.68 
Solve parts b and d of Problem 2.67 assuming that the free end of the 
rope is attached to the crate. 
Problem 2.67: A 280-kg crate is supported by several rope-and-pulley 
arrangements as shown. Determine for each arrangement the tension in 
the rope. (Hint: The tension in the rope is the same on each side of a 
simple pulley. This can be proved by the methods of Chapter 4.) 
 
SOLUTION 
Free-Body Diagram of pulley 
and crate 
(b) 
 
(d) 
 
 
 
 
 ( )( )20: 3 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
3
T = 
 916 NT = W 
 
 
 ( )( )20: 4 280 kg 9.81 m/s 0yF TΣ = − = 
( )1 2746.8 N
4
T = 
 687 NT = W 
 
 
 
 
 
 
 
 
 
 
 
 
 
70
 
 
PROBLEM 2.69 
A 350-lb load is supported by the rope-and-pulley arrangement shown. 
Knowing that β = 25°, determine the magnitude and direction of the 
force P which should be exerted on the free end of the rope to maintain 
equilibrium. (Hint: The tension in the rope is the same on each side of a 
simple pulley. This can be proved by the methods of Chapter 4.) 
 
SOLUTION 
Free-Body Diagram: Pulley A 
 
 
 0: 2 sin 25 cos 0xF P P αΣ = ° − = 
and 
cos 0.8452 or 32.3α α= = ± ° 
For 32.3α = + ° 
 0: 2 cos 25 sin 32.3 350 lb 0yF P PΣ = ° + ° − = 
 or 149.1 lb=P 32.3°W 
For 32.3α = − ° 
 0: 2 cos 25 sin 32.3 350 lb 0yF P PΣ = ° + − ° − = 
 or 274 lb=P 32.3°W 
 
71
 
PROBLEM 2.70 
A 350-lb load is supported by the rope-and-pulley arrangement shown. 
Knowing that 35 ,α = ° determine (a) the angle β, (b) the magnitude of 
the force P which should be exerted on the free end of the rope to 
maintain equilibrium. (Hint: The tension in the rope is the same on each 
side of a simple pulley. This can be proved by the methods of Chapter 4.) 
 
SOLUTION 
Free-Body Diagram: Pulley A 
 
 
 0: 2 sin cos 25 0xF P PβΣ = − ° = 
Hence: 
(a) 1sin cos 25
2
β = ° or 24.2β = °W 
(b) 0: 2 cos sin 35 350 lb 0yF P PβΣ = + ° − = 
 Hence: 
2 cos 24.2 sin 35 350 lb 0P P° + ° − = 
 or 145.97 lbP = 146.0 lbP = W 
 
 
72
 
 
 
 
PROBLEM 2.71 
A load Q is applied to the pulley C, which can roll on the cable ACB. The 
pulley is held in the position shown
by a second cable CAD, which passes 
over the pulley A and supports a load P. Knowing that P = 800 N, 
determine (a) the tension in cable ACB, (b) the magnitude of load Q. 
 
SOLUTION 
Free-Body Diagram: Pulley C 
 
(a) ( ) ( )0: cos30 cos50 800 N cos50 0x ACBF TΣ = ° − ° − ° = 
 Hence 2303.5 NACBT = 
 2.30 kN=ACBT W 
(b) ( ) ( )0: sin 30 sin 50 800 N sin 50 0y ACBF T QΣ = ° + ° + ° − = 
 ( )( ) ( )2303.5 N sin 30 sin 50 800 N sin 50 0Q° + ° + ° − = 
 or 3529.2 NQ = 3.53 kN=Q W 
73
 
 
 
 
PROBLEM 2.72 
A 2000-N load Q is applied to the pulley C, which can roll on the cable 
ACB. The pulley is held in the position shown by a second cable CAD, 
which passes over the pulley A and supports a load P. Determine (a) the 
tension in the cable ACB, (b) the magnitude of load P. 
 
SOLUTION 
Free-Body Diagram: Pulley C 
 
 ( )0: cos30 cos50 cos50 0x ACBF T PΣ = ° − ° − ° = 
or 0.3473 ACBP T= (1) 
 ( )0: sin 30 sin 50 sin 50 2000 N 0y ACBF T PΣ = ° + ° + ° − = 
or 1.266 0.766 2000 NACBT P+ = (2) 
(a) Substitute Equation (1) into Equation (2): 
 ( )1.266 0.766 0.3473 2000 NACB ACBT T+ = 
 Hence: 1305.5 NACBT = 
 1306 NACBT = W 
(b) Using (1) 
 ( )0.3473 1306 N 453.57 NP = = 
 454 NP = W 
74
 
 
 
 
PROBLEM 2.73 
Determine (a) the x, y, and z components of the 200-lb force, (b) the 
angles θx, θy, and θz that the force forms with the coordinate axes. 
 
SOLUTION 
(a) ( )200 lb cos30 cos 25 156.98 lbxF = ° ° = 
 157.0 lbxF = + W 
( )200 lb sin 30 100.0 lbyF = ° = 
 100.0 lbyF = + W 
( )200 lb cos30 sin 25 73.1996 lbzF = − ° ° = − 
 73.2 lbzF = − W 
(b) 156.98cos
200x
θ = or 38.3xθ = °W 
 100.0cos
200y
θ = or 60.0yθ = °W 
 73.1996cos
200z
θ −= or 111.5zθ = °W 
75
 
 
 
 
PROBLEM 2.74 
Determine (a) the x, y, and z components of the 420-lb force, (b) the 
angles θx, θy, and θz that the force forms with the coordinate axes. 
 
SOLUTION 
(a) ( )420 lb sin 20 sin 70 134.985 lbxF = − ° ° = − 
 135.0 lbxF = − W 
( )420 lb cos 20 394.67 lbyF = ° = 
 395 lbyF = + W 
( )420 lb sin 20 cos70 49.131 lbzF = ° ° = 
 49.1 lbzF = + W 
(b) 134.985cos
420x
θ −= 
 108.7xθ = °W 
394.67cos
420y
θ = 
 20.0yθ = °W 
49.131cos
420z
θ = 
 83.3zθ = °W 
76
 
 
 
 
PROBLEM 2.75 
To stabilize a tree partially uprooted in a storm, cables AB and AC are 
attached to the upper trunk of the tree and then are fastened to steel rods 
anchored in the ground. Knowing that the tension in cable AB is 4.2 kN, 
determine (a) the components of the force exerted by this cable on the 
tree, (b) the angles θx, θy, and θz that the force forms with axes at A which 
are parallel to the coordinate axes. 
 
SOLUTION 
 
(a) ( )4.2 kN sin 50 cos 40 2.4647 kNxF = ° ° = 
 2.46 kNxF = + W 
 ( )4.2 kN cos50 2.6997 kNyF = − ° = − 
 2.70 kNyF = − W 
 ( )4.2 kN sin 50 sin 40 2.0681 kNzF = ° ° = 
 2.07 kNzF = + W 
(b) 2.4647cos
4.2x
θ = 
 54.1xθ = °W 
77
 
PROBLEM 2.75 CONTINUED 
 2.7cos
4.2y
θ −= 
 130.0yθ = °W 
 2.0681cos
4.0z
θ = 
 60.5zθ = °W 
 
78
 
 
 
PROBLEM 2.76 
To stabilize a tree partially uprooted in a storm, cables AB and AC are 
attached to the upper trunk of the tree and then are fastened to steel rods 
anchored in the ground. Knowing that the tension in cable AC is 3.6 kN, 
determine (a) the components of the force exerted by this cable on the 
tree, (b) the angles θx, θy, and θz that the force forms with axes at A which 
are parallel to the coordinate axes. 
 
SOLUTION 
 
(a) ( )3.6 kN cos 45 sin 25 1.0758 kNxF = − ° ° = − 
 1.076 kNxF = − W 
( )3.6 kN sin 45 2.546 kNyF = − ° = − 
 2.55 kNyF = − W 
( )3.6 kN cos 45 cos 25 2.3071 kNzF = ° ° = 
 2.31 kNzF = + W 
(b) 1.0758cos
3.6x
θ −= 
 107.4xθ = °W 
79
 
PROBLEM 2.76 CONTINUED 
2.546cos
3.6y
θ −= 
 135.0yθ = °W 
2.3071cos
3.6z
θ = 
 50.1zθ = °W 
 
80
 
 
 
 
 
PROBLEM 2.77 
A horizontal circular plate is suspended as shown from three wires which 
are attached to a support at D and form 30° angles with the vertical. 
Knowing that the x component of the force exerted by wire AD on the 
plate is 220.6 N, determine (a) the tension in wire AD, (b) the angles θx, 
θy, and θz that the force exerted at A forms with the coordinate axes. 
 
SOLUTION 
(a) sin 30 sin 50 220.6 NxF F= ° ° = (Given) 
220.6 N 575.95 N
sin30 sin50
= =° °F 
 576 N=F W 
(b) 220.6cos 0.3830
575.95
θ = = =xx FF 
 67.5xθ = °W 
cos30 498.79 NyF F= ° = 
498.79cos 0.86605
575.95
y
y
F
F
θ = = = 
 30.0yθ = °W 
 sin 30 cos50zF F= − ° ° 
 ( )575.95 N sin 30 cos50= − ° ° 
 185.107 N= − 
185.107cos 0.32139
575.95
z
z
F
F
θ −= = = − 
 108.7zθ = °W 
81
 
 
 
 
PROBLEM 2.78 
A horizontal circular plate is suspended as shown from three wires which 
are attached to a support at D and form 30° angles with the vertical. 
Knowing that the z component of the force exerted by wire BD on the 
plate is –64.28 N, determine (a) the tension in wire BD, (b) the angles θx, 
θy, and θz that the force exerted at B forms with the coordinate axes. 
 
SOLUTION 
(a) sin 30 sin 40 64.28 NzF F= − ° ° = − (Given) 
 64.28 N 200.0 N
sin30 sin40
= =° °F 200 NF = W 
(b) sin 30 cos 40xF F= − ° ° 
 ( )200.0 N sin 30 cos 40= − ° ° 
 76.604 N= − 
 76.604cos 0.38302
200.0
x
x
F
F
θ −= = = − 112.5xθ = °W 
 cos30 173.2 NyF F= ° = 
 173.2cos 0.866
200
y
y
F
F
θ = = = 30.0yθ = °W 
 64.28 NzF = − 
 64.28cos 0.3214
200
z
z
F
F
θ −= = = − 108.7zθ = °W 
82
 
 
 
 
 
PROBLEM 2.79 
A horizontal circular plate is suspended as shown from three wires which 
are attached to a support at D and form 30° angles with the vertical. 
Knowing that the tension in wire CD is 120 lb, determine (a) the 
components of the force exerted by this wire on the plate, (b) the angles 
θx, θy, and θz that the force forms with the coordinate axes. 
 
SOLUTION 
(a) ( )120 lb sin 30 cos60 30 lbxF = − ° ° = − 
 30.0 lbxF = − W 
( )120 lb cos30 103.92 lbyF = ° = 
 103.9 lb= +yF W 
( )120 lb sin 30 sin 60 51.96 lbzF = ° ° = 
 52.0 lbzF = + W 
(b) 30.0cos 0.25
120
x
x
F
F
θ −= = = − 
 104.5xθ = °W 
103.92cos 0.866
120
y
y
F
F
θ = = = 
 30.0yθ = °W 
51.96cos 0.433
120
z
z
F
F
θ = = = 
 64.3zθ = °W 
83
 
 
 
 
PROBLEM 2.80 
A horizontal circular plate is suspended as shown from three wires which 
are attached to a support at D and form 30° angles with the vertical. 
Knowing that the x component of the forces exerted by wire CD on the 
plate is –40 lb, determine (a) the tension in wire CD, (b) the angles θx, θy, 
and θz that the force exerted at C forms with the coordinate axes. 
 
SOLUTION 
(a) sin 30 cos60 40 lbxF F= − ° ° = − (Given) 
40 lb 160 lb
sin30 cos60
= =° °F 
 160.0 lbF = W 
(b) 40cos 0.25
160
x
x
F
F
θ −= = = − 
 104.5xθ = °W 
( )160 lb cos30 103.92 lbyF = ° = 
 103.92cos 0.866
160
y
y
F
F
θ = = = 
 30.0yθ = °W 
( )160 lb sin 30 sin 60 69.282 lbzF = ° ° = 
69.282cos 0.433
160
z
z
F
F
θ = = = 
 64.3zθ = °W 
 
84
 
 
 
 
 
PROBLEM 2.81 
Determine the magnitude and direction of the force 
( ) ( ) ( )800 lb 260 lb 320 lb .= + −F i j k 
 
SOLUTION 
 ( ) ( ) ( )2 2 22 2 2 800 lb 260 lb 320 lbx y zF F F F= + + = + + − 900 lbF = W 
 800cos 0.8889
900
x
x
F
F
θ = = = 27.3xθ = °W 
 260cos 0.2889
900
y
y
F
F
θ = = = 73.2yθ = °W 
 320cos
0.3555
900
z
z
F
F
θ −= = = − 110.8zθ = °W 
85
 
 
 
 
 
PROBLEM 2.82 
Determine the magnitude and direction of the force 
( ) ( ) ( )400 N 1200 N 300 N .= − +F i j k 
 
SOLUTION 
 ( ) ( ) ( )2 2 22 2 2 400 N 1200 N 300 Nx y zF F F F= + + = + − + 1300 NF = W 
 400cos 0.30769
1300
x
x
F
F
θ = = = 72.1xθ = °W 
 1200cos 0.92307
1300
y
y
F
F
θ −= = = − 157.4yθ = °W 
 300cos 0.23076
1300
z
z
F
F
θ = = = 76.7zθ = °W 
86
 
 
 
 
 
 
PROBLEM 2.83 
A force acts at the origin of a coordinate system in a direction defined by 
the angles θx = 64.5° and θz = 55.9°. Knowing that the y component of 
the force is –200 N, determine (a) the angle θy, (b) the other components 
and the magnitude of the force. 
 
SOLUTION 
(a) We have 
( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2cos cos cos 1 cos 1 cos cosx y z y y zθ θ θ θ θ θ+ + = ⇒ = − − 
 Since 0yF < we must have cos 0yθ < 
 Thus, taking the negative square root, from above, we have: 
 ( ) ( )2 2cos 1 cos64.5 cos55.9 0.70735yθ = − − ° − ° = − 135.0yθ = °W 
(b) Then: 
 200 N 282.73 N
cos 0.70735
y
y
F
F θ
−= = =− 
 and ( )cos 282.73 N cos64.5x xF F θ= = ° 121.7 NxF = W 
 ( )cos 282.73 N cos55.9z zF F θ= = ° 158.5 NyF = W 
 283 NF = W 
87
 
 
 
 
 
PROBLEM 2.84 
A force acts at the origin of a coordinate system in a direction defined by 
the angles θx = 75.4° and θy = 132.6°. Knowing that the z component of 
the force is –60 N, determine (a) the angle θz, (b) the other components 
and the magnitude of the force. 
 
SOLUTION 
(a) We have 
( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2cos cos cos 1 cos 1 cos cosx y z y y zθ θ θ θ θ θ+ + = ⇒ = − − 
 Since 0zF < we must have cos 0zθ < 
 Thus, taking the negative square root, from above, we have: 
 ( ) ( )2 2cos 1 cos75.4 cos132.6 0.69159zθ = − − ° − ° = − 133.8zθ = °W 
(b) Then: 
 60 N 86.757 N
cos 0.69159
z
z
FF θ
−= = =− 86.8 NF = W 
 and ( )cos 86.8 N cos75.4x xF F θ= = ° 21.9 NxF = W 
 ( )cos 86.8 N cos132.6y yF F θ= = ° 58.8 NyF = − W 
88
 
 
 
 
 
 
PROBLEM 2.85 
A force F of magnitude 400 N acts at the origin of a coordinate system. 
Knowing that θx = 28.5°, Fy = –80 N, and Fz > 0, determine (a) the 
components Fx and Fz, (b) the angles θy and θz. 
 
SOLUTION 
(a) Have 
 ( )cos 400 N cos 28.5x xF F θ= = ° 351.5 NxF = W 
 Then: 
2 2 2 2
x y zF F F F= + + 
 So: ( ) ( ) ( )2 2 2 2400 N 352.5 N 80 N zF= + − + 
 Hence: 
 ( ) ( ) ( )2 2 2400 N 351.5 N 80 NzF = + − − − 173.3 NzF = W 
(b) 
 80cos 0.20
400
y
y
F
F
θ −= = = − 101.5yθ = °W 
 173.3cos 0.43325
400
z
z
F
F
θ = = = 64.3zθ = °W 
89
 
 
 
 
 
PROBLEM 2.86 
A force F of magnitude 600 lb acts at the origin of a coordinate system. 
Knowing that Fx = 200 lb, θz = 136.8°, Fy < 0, determine (a) the 
components Fy and Fz, (b) the angles θx and θy. 
 
SOLUTION 
(a) ( )cos 600 lb cos136.8z zF F θ= = ° 
 437.4 lb= − 437 lbzF = − W 
 Then: 
2 2 2 2
x y zF F F F= + + 
 So: ( ) ( ) ( ) ( )22 2 2600 lb 200 lb 437.4 lbyF= + + − 
 Hence: ( ) ( ) ( )2 2 2600 lb 200 lb 437.4 lbyF = − − − − 
 358.7 lb= − 359 lbyF = − W 
(b) 
 200cos 0.333
600
x
x
F
F
θ = = = 70.5xθ = °W 
 358.7cos 0.59783
600
y
y
F
F
θ −= = = − 126.7yθ = °W 
90
 
 
 
 
 
PROBLEM 2.87 
A transmission tower is held by three guy wires anchored by bolts at B, 
C, and D. If the tension in wire AB is 2100 N, determine the components 
of the force exerted by the wire on the bolt at B. 
 
SOLUTION 
( ) ( ) ( )4 m 20 m 5 mBA = + −i j kJJJG 
( ) ( ) ( )2 2 24 m 20 m 5 m 21 mBA = + + − = 
( ) ( ) ( )2100 N 4 m 20 m 5 m
21 mBA
BAF F
BA
 = = = + − F i j k
JJJG
λ 
( ) ( ) ( )400 N 2000 N 500 N= + −F i j k 
 400 N, 2000 N, 500 Nx y zF F F= + = + = − W 
91
 
 
 
92
 
PROBLEM 2.88 
A transmission tower is held by three guy wires anchored by bolts at B, 
C, and D. If the tension in wire AD is 1260 N, determine the components 
of the force exerted by the wire on the bolt at D. 
 
SOLUTION 
( ) ( ) ( )4 m 20 m 14.8 mDA = + +i j kJJJG 
( ) ( ) ( )2 2 24 m 20 m 14.8 m 25.2 mDA = + + = 
( ) ( ) ( )1260 N 4 m 20 m 14.8 m
25.2 mDA
DAF F
DA
 = = = + + F i j k
JJJG
λ 
( ) ( ) ( )200 N 1000 N 740 N= + +F i j k 
 200 N, 1000 N, 740 Nx y zF F F= + = + = + W 
 
 
 
 
 
PROBLEM 2.89 
A rectangular plate is supported by three cables as shown. Knowing that 
the tension in cable AB is 204 lb, determine the components of the force 
exerted on the plate at B. 
 
SOLUTION 
( ) ( ) ( )32 in. 48 in. 36 in.BA = + −i j kJJJG 
 ( ) ( ) ( )2 2 232 in. 48 in. 36 in. 68 in.BA = + + − = 
( ) ( ) ( )204 lb 32 in. 48 in. 36 in.
68 in.BA
BAF F
BA
 = = = + − F i j k
JJJG
λ 
( ) ( ) ( )96 lb 144 lb 108 lb= + −F i j k 
 96.0 lb, 144.0 lb, 108.0 lbx y zF F F= + = + = − W 
93
 
 
 
 
PROBLEM 2.90 
A rectangular plate is supported by three cables as shown. Knowing that 
the tension in cable AD is 195 lb, determine the components of the force 
exerted on the plate at D. 
 
SOLUTION 
( ) ( ) ( )25 in. 48 in. 36 in.DA = − + +i j kJJJG 
 ( ) ( ) ( )2 2 225 in. 48 in. 36 in. 65 in.DA = − + + = 
( ) ( ) ( )195 lb 25 in. 48 in. 36 in.
65 in.DA
DAF F
DA
 = = = − + + F i j k
JJJG
λ 
( ) ( ) ( )75 lb 144 lb 108 lb= − + +F i j k 
 75.0 lb, 144.0 lb, 108.0 lbx y zF F F= − = + = + W 
 
94
 
 
 
 
PROBLEM 2.91 
A steel rod is bent into a semicircular ring of radius 0.96 m and is 
supported in part by cables BD and BE which are attached to the ring at 
B. Knowing that the tension in cable BD is 220 N, determine the 
components of this force exerted by the cable on the support at D. 
 
SOLUTION 
( ) ( ) ( )0.96 m 1.12 m 0.96 mDB = − −i j kJJJG 
( ) ( ) ( )2 2 20.96 m 1.12 m 0.96 m 1.76 mDB = + − + − = 
( ) ( ) ( )220 N 0.96 m 1.12 m 0.96 m
1.76 mDB DB
DBT T
DB
 = = = − − T i j k
JJJG
λ 
( ) ( ) ( )120 N 140 N 120 NDB = − −T i j k 
 ( ) ( ) ( )120.0 N, 140.0 N, 120.0 NDB DB DBx y zT T T= + = − = − W 
95
 
 
 
 
PROBLEM 2.92 
A steel rod is bent into a semicircular ring of radius 0.96 m and is 
supported in part by cables BD and BE which are attached to the ring at 
B. Knowing that the tension in cable BE is 250 N, determine the 
components of this force exerted by the cable on the support at E. 
 
SOLUTION 
( ) ( ) ( )0.96 m 1.20 m 1.28 mEB = − +i j kJJJG 
( ) ( ) ( )2 2 20.96 m 1.20 m 1.28 m 2.00 mEB = + − + = 
( ) ( ) ( )250 N 0.96 m 1.20 m 1.28 m
2.00 mEB EB
EBT T
EB
 = = = − + T i j k
JJJG
λ 
( ) ( ) ( )120 N 150 N 160 NEB = − +T i j k 
 ( ) ( ) ( )120.0 N, 150.0 N, 160.0 NEB EB EBx y zT T T= + = − = + W 
96
 
 
 
 
 
PROBLEM 2.93 
Find the magnitude and direction of the resultant of the two forces shown 
knowing that 500 NP = and 600 N.Q = 
 
SOLUTION 
( )[ ]500 lb cos30 sin15 sin 30 cos30 cos15= − ° ° + ° + ° °P i j k 
 ( )[ ]500 lb 0.2241 0.50 0.8365= − + +i j k 
 ( ) ( ) ( )112.05 lb 250 lb 418.25 lb= − + +i j k 
( )[ ]600 lb cos 40 cos 20 sin 40 cos 40 sin 20= ° ° + ° − ° °Q i j k 
 ( )[ ]600 lb 0.71985 0.64278 0.26201= + −i j k 
 ( ) ( ) ( )431.91 lb 385.67 lb 157.206 lb= + −i j k 
( ) ( ) ( )319.86 lb 635.67 lb 261.04 lb= + = + +R P Q i j k 
( ) ( ) ( )2 2 2319.86 lb 635.67 lb 261.04 lb 757.98 lbR = + + = 
 758 lbR = W 
319.86 lbcos 0.42199
757.98 lb
x
x
R
R
θ = = = 
 65.0xθ = °W 
635.67 lbcos 0.83864
757.98 lb
y
y
R
R
θ = = = 
 33.0yθ = °W 
261.04 lbcos 0.34439
757.98 lb
z
z
R
R
θ = = =
69.9zθ = °W 
97
 
 
 
 
PROBLEM 2.94 
Find the magnitude and direction of the resultant of the two forces shown 
knowing that P = 600 N and Q = 400 N. 
 
SOLUTION 
Using the results from 2.93: 
( )[ ]600 lb 0.2241 0.50 0.8365= − + +P i j k 
 ( ) ( ) ( )134.46 lb 300 lb 501.9 lb= − + +i j k 
( )[ ]400 lb 0.71985 0.64278 0.26201= + −Q i j k 
 ( ) ( ) ( )287.94 lb 257.11 lb 104.804 lb= + −i j k 
( ) ( ) ( )153.48 lb 557.11 lb 397.10 lb= + = + +R P Q i j k 
( ) ( ) ( )2 2 2153.48 lb 557.11 lb 397.10 lb 701.15 lbR = + + = 
 701 lbR = W 
153.48 lbcos 0.21890
701.15 lb
x
x
R
R
θ = = = 
 77.4xθ = °W 
557.11 lbcos 0.79457
701.15 lb
y
y
R
R
θ = = = 
 37.4yθ = °W 
397.10 lbcos 0.56637
701.15 lb
z
z
R
R
θ = = = 
 55.5zθ = °W 
98
 
 
 
 
PROBLEM 2.95 
Knowing that the tension is 850 N in cable AB and 1020 N in cable AC, 
determine the magnitude and direction of the resultant of the forces 
exerted at A by the two cables. 
 
SOLUTION 
( ) ( ) ( )400 mm 450 mm 600 mmAB = − +i j kJJJG 
( ) ( ) ( )2 2 2400 mm 450 mm 600 mm 850 mmAB = + − + = 
( ) ( ) ( )1000 mm 450 mm 600 mmAC = − +i j kJJJG 
( ) ( ) ( )2 2 21000 mm 450 mm 600 mm 1250 mmAC = + − + = 
( ) ( ) ( ) ( )400 mm 450 mm 600 mm850 N
850 mmAB AB ABAB
ABT T
AB
 − += = =   
i j k
T
JJJG
λ 
( ) ( ) ( )400 N 450 N 600 NAB = − +T i j k 
( ) ( ) ( ) ( )1000 mm 450 mm 600 mm1020 N
1250 mmAC AC ACAC
ACT T
AC
 − += = =   
i j k
T
JJJG
λ 
( ) ( ) ( )816 N 367.2 N 489.6 NAC = − +T i j k 
( ) ( ) ( )1216 N 817.2 N 1089.6 NAB AC= + = − +R T T i j k 
Then: 1825.8 NR = 1826 NR = W 
and 1216cos 0.66601
1825.8x
θ = = 48.2xθ = °W 
 817.2cos 0.44758
1825.8y
θ −= = − 116.6yθ = °W 
 1089.6cos 0.59678
1825.8z
θ = = 53.4zθ = °W 
99
 
 
 
PROBLEM 2.96 
Assuming that in Problem 2.95 the tension is 1020 N in cable AB and 
850 N in cable AC, determine the magnitude and direction of the resultant 
of the forces exerted at A by the two cables. 
 
SOLUTION 
( ) ( ) ( )400 mm 450 mm 600 mmAB = − +i j kJJJG 
( ) ( ) ( )2 2 2400 mm 450 mm 600 mm 850 mmAB = + − + = 
( ) ( ) ( )1000 mm 450 mm 600 mmAC = − +i j kJJJG 
( ) ( ) ( )2 2 21000 mm 450 mm 600 mm 1250 mmAC = + − + = 
( ) ( ) ( ) ( )400 mm 450 mm 600 mm1020 N
850 mmAB AB AB AB
ABT T
AB
 − += = =   
i j k
T
JJJG
λ 
( ) ( ) ( )480 N 540 N 720 NAB = − +T i j k 
( ) ( ) ( ) ( )1000 mm 450 mm 600 mm850 N
1250 mmAC AC AC AC
ACT T
AC
 − += = =   
i j k
T
JJJG
λ 
( ) ( ) ( )680 N 306 N 408 NAC = − +T i j k 
( ) ( ) ( )1160 N 846 N 1128 NAB AC= + = − +R T T i j k 
Then: 1825.8 NR = 1826 NR = W 
and 1160cos 0.6353
1825.8x
θ = = 50.6xθ = °W 
 846cos 0.4634
1825.8y
θ −= = − 117.6yθ = °W 
 1128cos 0.6178
1825.8z
θ = = 51.8zθ = °W 
100
 
 
 
 
PROBLEM 2.97 
For the semicircular ring of Problem 2.91, determine the magnitude and 
direction of the resultant of the forces exerted by the cables at B knowing 
that the tensions in cables BD and BE are 220 N and 250 N, respectively. 
 
SOLUTION 
For the solutions to Problems 2.91 and 2.92, we have 
( ) ( ) ( )120 N 140 N 120 NBD = − + +T i j k 
( ) ( ) ( )120 N 150 N 160 NBE = − + −T i j k 
Then: 
 B BD BE= +R T T 
 ( ) ( ) ( )240 N 290 N 40 N= − + −i j k 
and 378.55 NR = 379 NBR = W 
240cos 0.6340
378.55x
θ = − = − 
 129.3xθ = °W 
290cos 0.7661
378.55y
θ = = − 
 40.0yθ = °W 
40cos 0.1057
378.55z
θ = − = − 
 96.1zθ = °W 
 
101
 
PROBLEM 2.98 
To stabilize a tree partially uprooted in a storm, cables AB and AC are 
attached to the upper trunk of the tree and then are fastened to steel rods 
anchored in the ground. Knowing that the tension in AB is 920 lb and that 
the resultant of the forces exerted at A by cables AB and AC lies in the yz 
plane, determine (a) the tension in AC, (b) the magnitude and direction of 
the resultant of the two forces. 
 
SOLUTION 
Have 
( )( )920 lb sin 50 cos 40 cos50 sin 50 sin 40AB = ° ° − ° + ° °T i j j 
( )cos 45 sin 25 sin 45 cos 45 cos 25AC ACT= − ° ° − ° + ° °T i j j 
(a) 
A AB AC= +R T T 
( ) 0A xR = 
∴ ( ) ( )0: 920 lb sin 50 cos 40 cos 45 sin 25 0A x ACxR F T= Σ = ° ° − ° ° = 
 or 
 1806.60 lbACT = 1807 lbACT = W 
(b) 
( ) ( ) ( ): 920 lb cos50 1806.60 lb sin 45A yyR F= Σ − ° − ° 
( ) 1868.82 lbA yR = − 
( ) ( ) ( ): 920 lb sin 50 sin 40 1806.60 lb cos 45 cos 25A zzR F= Σ ° ° + ° ° 
( ) 1610.78 lbA zR = 
∴ ( ) ( )1868.82 lb 1610.78 lbAR = − +j k 
 Then: 
 2467.2 lbAR = 2.47 kipsAR = W 
102
 
 
PROBLEM 2.98 CONTINUED 
 and 
 0cos 0
2467.2x
θ = = 90.0xθ = °W 
 1868.82cos 0.7560
2467.2y
θ −= = − 139.2yθ = °W 
 1610.78cos 0.65288
2467.2z
θ = = 49.2zθ = °W 
 
103
 
PROBLEM 2.99 
To stabilize a tree partially uprooted in a storm, cables AB and AC are 
attached to the upper trunk of the tree and then are fastened to steel rods 
anchored in the ground. Knowing that the tension in AC is 850 lb and that 
the resultant of the forces exerted at A by cables AB and AC lies in the yz 
plane, determine (a) the tension in AB, (b) the magnitude and direction of 
the resultant of the two forces. 
 
SOLUTION 
Have 
( )sin 50 cos 40 cos50 sin 50 sin 40AB ABT= ° ° − ° + ° °T i j j 
( )( )850 lb cos 45 sin 25 sin 45 cos 45 cos 25AC = − ° ° − ° + ° °T i j j 
(a) 
( ) 0A xR = 
∴ ( ) ( )0: sin 50 cos 40 850 lb cos 45 sin 25 0A x ABxR F T= Σ = ° ° − ° ° = 
 432.86 lbABT = 433 lbABT = W 
(b) 
( ) ( ) ( ): 432.86 lb cos50 850 lb sin 45A yyR F= Σ − ° − ° 
( ) 879.28 lbA yR = − 
( ) ( ) ( ): 432.86 lb sin 50 sin 40 850 lb cos 45 cos 25A zzR F= Σ ° ° + ° ° 
( ) 757.87 lbA zR = 
∴ ( ) ( )879.28 lb 757.87 lbA = − +R j k 
 1160.82 lbAR = 1.161 kipsAR = W 
 0cos 0
1160.82x
θ = = 90.0xθ = °W 
 879.28cos 0.75746
1160.82y
θ −= = − 139.2yθ = °W 
 757.87cos 0.65287
1160.82z
θ = = 49.2zθ = °W 
 
104
 
 
PROBLEM 2.100 
For the plate of Problem 2.89, determine the tension in cables AB and AD 
knowing that the tension if cable AC is 27 lb and that the resultant of the 
forces exerted by the three cables at A must be vertical. 
 
SOLUTION 
With: 
( ) ( ) ( )45 in. 48 in. 36 in.AC = − +i j kJJJG 
( ) ( ) ( )2 2 245 in. 48 in. 36 in. 75 in.AC = + − + = 
( ) ( ) ( )27 lb 45 in. 48 in. 36 in.
75 in.AC AC AC AC
ACT T
AC
 = = = − + T i j k
JJJG
λ 
( ) ( ) ( )16.2 lb 17.28 lb 12.96AC = − +T i j k 
and 
( ) ( ) ( )32 in. 48 in. 36 in.AB = − − +i j kJJJG 
( ) ( ) ( )2 2 232 in. 48 in. 36 in. 68 in.AB = − + − + = 
( ) ( ) ( )32 in. 48 in. 36 in.
68 in.
AB
AB AB AB AB
AB TT T
AB
 = = = − − + T i j k
JJJG
λ 
( )0.4706 0.7059 0.5294AB ABT= − − +T i j k 
and 
( ) ( ) ( )25 in. 48 in. 36 in.AD = − −i j kJJJG 
( ) ( ) ( )2 2 225 in. 48 in. 36 in. 65 in.AD = + − + = 
( ) ( ) ( )25 in. 48 in. 36 in.
65 in.
AD
AD AD AD AD
AD TT T
AD
 = = = − − T i j k
JJJG
λ 
( )0.3846 0.7385 0.5538AD ADT= − −T i j k 
 
105
 
PROBLEM 2.100 CONTINUED 
Now 
 AB AD AD= + +R T T T 
 ( ) ( ) ( ) ( )0.4706 0.7059 0.5294 16.2 lb 17.28 lb 12.96ABT  = − − + + − + i j k i j k 
 ( )0.3846 0.7385 0.5538ADT+ − −i j k 
Since R must be vertical, the i and k components of this sum must be zero. 
Hence: 
 0.4706 0.3846 16.2 lb 0AB ADT T− + + = (1) 
 0.5294 0.5538 12.96 lb 0AB ADT T− + = (2) 
Solving (1) and (2), we obtain: 
244.79 lb, 257.41 lbAB ADT T= = 
 245 lbABT = W 
 257 lbADT = W 
 
106
 
 
 
 
 
PROBLEM 2.101 
The support assembly shown is bolted in place at B, C, and D and 
supports a downward force P at A. Knowing that the forces
in members 
AB, AC, and AD are directed along the respective members and that the 
force in member AB is 146 N, determine the magnitude of P. 
 
SOLUTION 
Note that AB, AC, and AD are in compression. 
Have 
( ) ( ) ( )2 2 2220 mm 192 mm 0 292 mmBAd = − + + = 
 ( ) ( ) ( )2 2 2192 mm 192 mm 96 mm 288 mmDAd = + + = 
 ( ) ( ) ( )2 2 20 192 mm 144 mm 240 mmCAd = + + − = 
and ( ) ( )146 N 220 mm 192 mm
292 mmBA BA BA
F  = = − + F i jλ 
 ( ) ( )110 N 96 N= − +i j 
( ) ( )192 mm 144 mm
240 mm
CA
CA CA CA
FF  = = − F j kλ 
 ( )0.80 0.60CAF= −j k 
( ) ( ) ( )192 mm 192 mm 96 mm
288 mm
DA
DA DA DA
FF  = = + + F i j kλ 
 [ ]0.66667 0.66667 0.33333DAF= + +i j k 
With P= −P j 
At A: 0: 0BA CA DAΣ = + + + =F F F F P 
i-component: ( )110 N 0.66667 0DAF− + = or 165 NDAF = 
j-component: ( )96 N 0.80 0.66667 165 N 0CAF P+ + − = (1) 
k-component: ( )0.60 0.33333 165 N 0CAF− + = (2) 
Solving (2) for CAF and then using that result in (1), gives 279 NP = W 
107
 
 
 
 
 
PROBLEM 2.102 
The support assembly shown is bolted in place at B, C, and D and 
supports a downward force P at A. Knowing that the forces in members 
AB, AC, and AD are directed along the respective members and that 
P = 200 N, determine the forces in the members. 
 
SOLUTION 
With the results of 2.101: 
 ( ) ( )220 mm 192 mm
292 mm
BA
BA BA BA
FF  = = − + F i jλ 
 [ ]0.75342 0.65753 NBAF= − +i j 
( ) ( )192 mm 144 mm
240 mm
CA
CA CA CA
FF  = = − F j kλ 
 ( )0.80 0.60CAF= −j k 
( ) ( ) ( )192 mm 192 mm 96 mm
288 mm
DA
DA DA DA
FF  = = + + F i j kλ 
 [ ]0.66667 0.66667 0.33333DAF= + +i j k 
With: ( )200 N= −P j 
At A: 0: 0BA CA DAΣ = + + + =F F F F P 
Hence, equating the three (i, j, k) components to 0 gives three equations 
i-component: 0.75342 0.66667 0BA DAF F− + = (1) 
j-component: 0.65735 0.80 0.66667 200 N 0BA CA DAF F F+ + − = (2) 
k-component: 0.60 0.33333 0CA DAF F− + = (3) 
Solving (1), (2), and (3), gives 
 DA104.5 N, 65.6 N, 118.1 NBA CAF F F= = = 
 104.5 NBAF = W 
 65.6 NCAF = W 
 118.1 NDAF = W 
108
 
 
 
 
 
PROBLEM 2.103 
Three cables are used to tether a balloon as shown. Determine the vertical 
force P exerted by the balloon at A knowing that the tension in cable AB 
is 60 lb. 
 
SOLUTION 
 
 
 
The forces applied at A are: 
, , and AB AC ADT T T P 
where P=P j . To express the other forces in terms of the unit vectors 
i, j, k, we write 
 ( ) ( )12.6 ft 16.8 ftAB = − −i jJJJG 21 ftAB = 
( ) ( ) ( )7.2 ft 16.8 ft 12.6 ft 22.2 ftAC AC= − + =i j kJJJG 
 ( ) ( )16.8 ft 9.9 ftAD = − −j kJJJG 19.5 ftAD = 
and ( )0.6 0.8AB AB AB AB ABABT T TAB= = = − −T i j
JJJG
λ 
( )0.3242 0.75676 0.56757AC AC AC AC ACACT T TAC= = = − +T i j k
JJJG
λ 
( )0.8615 0.50769AD AD AD AD ADADT T TAD= = = − −T j k
JJJG
λ 
109
 
 
 
 
 
PROBLEM 2.103 CONTINUED 
Equilibrium Condition 
0: 0AB AC ADF PΣ = + + + =T T T j 
Substituting the expressions obtained for , , and AB AC ADT T T and 
factoring i, j, and k: 
( ) ( )0.6 0.3242 0.8 0.75676 0.8615AB AC AB AC ADT T T T T P− + + − − − +i j 
 ( )0.56757 0.50769 0AC ADT T+ − =k 
Equating to zero the coefficients of i, j, k: 
 0.6 0.3242 0AB ACT T− + = (1) 
 0.8 0.75676 0.8615 0AB AC ADT T T P− − − + = (2) 
 0.56757 0.50769 0AC ADT T− = (3) 
Setting 60 lbABT = in (1) and (2), and solving the resulting set of 
equations gives 
111 lbACT = 
124.2 lbADT = 
 239 lb=P W 
 
110
 
 
 
 
 
PROBLEM 2.104 
Three cables are used to tether a balloon as shown. Determine the vertical 
force P exerted by the balloon at A knowing that the tension in cable AC 
is 100 lb. 
 
SOLUTION 
See Problem 2.103 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) 
below: 
 0.6 0.3242 0AB ACT T− + = (1) 
 0.8 0.75676 0.8615 0AB AC ADT T T P− − − + = (2) 
 0.56757 0.50769 0AC ADT T− = (3) 
Substituting 100 lbACT = in Equations (1), (2), and (3) above, and solving the resulting set of equations 
using conventional algorithms gives 
54 lbABT = 
112 lbADT = 
 215 lb=P W 
111
 
 
 
 
 
PROBLEM 2.105 
The crate shown in Figure P2.105 and P2.108 is supported by three 
cables. Determine the weight of the crate knowing that the tension in 
cable AB is 3 kN. 
 
SOLUTION 
 
 
 
The forces applied at A are: 
, , and AB AC ADT T T P 
where P=P j . To express the other forces in terms of the unit vectors 
i, j, k, we write 
 ( ) ( ) ( )0.72 m 1.2 m 0.54 m ,AB = − + −i j kJJJG 1.5 mAB = 
 ( ) ( )1.2 m 0.64 m ,AC = +j kJJJG 1.36 mAC = 
 ( ) ( ) ( )0.8 m 1.2 m 0.54 m ,AD = + −i j kJJJG 1.54 mAD = 
and ( )0.48 0.8 0.36AB AB AB AB ABABT T TAB= = = − + −T i j k
JJJG
λ 
( )0.88235 0.47059AC AC AC AC ACACT T TAC= = = +T j k
JJJG
λ 
( )0.51948 0.77922 0.35065AD AD AD AD ADADT T TAD= = = + −T i j k
JJJG
λ 
Equilibrium Condition with W= −W j 
0: 0AB AC ADF WΣ = + + − =T T T j 
Substituting the expressions obtained for , , and AB AC ADT T T and 
factoring i, j, and k: 
( ) ( )0.48 0.51948 0.8 0.88235 0.77922AB AD AB AC ADT T T T T W− + + + + −i j
 ( )0.36 0.47059 0.35065 0AB AC ADT T T+ − + − =k 
112
 
 
 
 
 
PROBLEM 2.105 CONTINUED 
Equating to zero the coefficients of i, j, k: 
0.48 0.51948 0AB ADT T− + = 
0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = 
 0.36 0.47059 0.35065 0AB AC ADT T T− + − = 
Substituting 3 kNABT = in Equations (1), (2) and (3) and solving the 
resulting set of equations, using conventional algorithms for solving 
linear algebraic equations, gives 
4.3605 kNACT = 
2.7720 kNADT = 
 8.41 kNW = W 
 
113
 
 
 
 
 
PROBLEM 2.106 
For the crate of Problem 2.105, determine the weight of the crate 
knowing that the tension in cable AD is 2.8 kN. 
Problem 2.105: The crate shown in Figure P2.105 and P2.108 is 
supported by three cables. Determine the weight of the crate knowing that 
the tension in cable AB is 3 kN. 
 
SOLUTION 
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) 
below: 
 0.48 0.51948 0AB ADT T− + = 
 0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = 
 0.36 0.47059 0.35065 0AB AC ADT T T− + − = 
Substituting 2.8 kNADT = in Equations (1), (2), and (3) above, and solving the resulting set of equations 
using conventional algorithms, gives 
3.03 kNABT = 
4.40 kNACT = 
 8.49 kNW = W 
114
 
 
 
 
PROBLEM 2.107 
For the crate of Problem 2.105, determine the weight of the crate 
knowing that the tension in cable AC is 2.4 kN. 
Problem 2.105: The crate shown in Figure P2.105 and P2.108 is 
supported by three cables. Determine the weight of the crate knowing that 
the tension in cable AB is 3 kN. 
 
SOLUTION 
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) 
below: 
 0.48 0.51948 0AB ADT T− + = 
 0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = 
 0.36 0.47059 0.35065 0AB AC ADT T T− + − = 
Substituting 2.4 kNACT = in Equations (1), (2), and (3) above, and solving the resulting set of equations 
using conventional algorithms, gives 
1.651 kNABT = 
1.526 kNADT = 
 4.63 kNW = W 
115
 
 
 
 
 
PROBLEM 2.108 
A 750-kg crate is supported by three cables as shown. Determine the 
tension in each cable. 
 
SOLUTION 
See Problem 2.105 for the figure and the analysis leading to the linear algebraic Equations (1), (2), and (3) 
below: 
 0.48 0.51948 0AB ADT T− + = 
 0.8 0.88235 0.77922 0AB AC ADT T T W+ + − = 
 0.36 0.47059 0.35065 0AB AC ADT